Question

Let $I_n={\int }_0^{\infty }{e}^{-x}(sinx{)}^{n}dx,nϵN,n>1$ then $\frac{I_{2008}}{I_{2006}}$ equals

• A. $\frac{2007\times 2006}{{2008}^2+1}$
• B. $\frac{2008\times 2007}{{2008}^2+1}$
• C. $\frac{2006\times 2004}{{2008}^2-1}$
• D. $\frac{2008\times 2007}{{2008}^2-1}$

Answer 1

The correct option is B $\frac{2008\times 2007}{{2008}^2+1}$

$I_n={\int }_0^{\infty }{e}^{-x}(sinx{)}^{n}dx=[si{n}^{n}x(-e^{-x}){]}_0^{\infty }+{\int }_0^{\infty }nsi{n}^{n-1}xcosx{e}^{-x}dx=0+n{\int }_0^{\infty }\left(si{n}^{n-1}xcosx\right)e^{-x}dx=n\left[\right(si{n}^{n-1}xcosx\left)\right(-e^{-x}){]}_0^{\infty }-n{\int }_0^{\infty }\{-si{n}^{n}x+(n-1\left)si{n}^{n-2}xco{s}^{2}x\right\}(-e^{-x})dx=0+n{\int }_0^{\infty }{e}^{-x}\left\{\right(n-1)si{n}^{2}x+(n-1\left)si{n}^{n-2}x\right(1-si{n}^{2}x\left)\right\}dx=n{\int }_0^{\infty }{e}^{-x}\left\{\right(n-1)si{n}^{n-2}x-nsi{n}^{n}x\}dx=n(n-1)I_{n-2}-n^2{I}_n$
we have $(1+n^2)I_n=n(n-1)I_{n-2}$
then $\frac{I_n}{I_n-2}=\frac{n(n-1)}{n^2+1}$